Consider the functions
$(1-a)x,\quad 0 ≤ x ≤ a,$
$a(1-x),\quad a ≤ x ≤ 1,$
over the interval $[0, 1]$ and $0 < a < 1$.
My question is: since this is not over a symmetric interval, how would I go about representing this function as a Fourier series? Any help would be appreciated!
If you want to expand with respect to the orthonormal system $U,C_1,S_1,C_2,S_2,\cdots$ with $U(x)=1$, $C_n(x)=\frac{\sqrt2}2\cos(2\pi nx)$ and $S_n(x)=\frac{\sqrt2}2\sin(2\pi nx)$, then recall $\int x\cos(bx)=\frac xb\sin(bx)+\frac1{b^2}\cos(bx)$ and $\int x\sin(bx)=-\frac xb\cos(bx)+\frac1{b^2}\sin(bx)$. Thus \begin{align}\int_0^1f(x)U(x)\,dx&=(1-a)\frac{a^2}2+a\left(1-\frac12-a+\frac{a^2}2\right)=-\frac{a^2+a}2\\ \int_0^1f(x)C_n(x)&=2^{-1/2}(1-a)\int_0^a(1-x)\cos(2\pi nx)\,dx+2^{-1/2}a\int_a^1x\cos(2\pi nx)\,dx=\\&=2^{-1/2}(1-a)\left(\frac{1}{2\pi n}\sin(2\pi na)-\frac a{2\pi n}\sin(2\pi na)-\frac1{4\pi^2n^2}\cos(2\pi na)+\frac1{4\pi^2n^2}\right)+\\&+2^{-1/2}a\left(\frac1{4\pi^2n^2}-\frac{a}{2\pi n}\sin(2\pi na)-\frac1{4\pi^2n^2}\cos(2\pi na)\right)=\\&=2^{-1/2}\left(\frac{1-\cos(2\pi na)}{4\pi^2n^2}+\frac{(1-2a)}{2\pi n}\sin(2\pi na)\right)\\\int_0^1f(x)S_n(x)&=2^{-1/2}(1-a)\int_0^a(1-x)\sin(2\pi nx)\,dx+2^{-1/2}a\int_a^1x\sin(2\pi nx)\,dx=\\&=2^{-1/2}(1-a)\left(\frac{1-\cos(2\pi na)}{2\pi n}+\frac{a}{2\pi n}\cos(2\pi na)-\frac1{4\pi^2n^2}\sin(2\pi na)\right)+\\&+2^{-1/2}a\left(-\frac{1}{2\pi n}+\frac a{2\pi n}\cos(2\pi na)-\frac1{4\pi^2n^2}\sin(2\pi na)\right)=\\&=2^{-1/2}\left(\frac{(1-2a)(1+\cos(2\pi na))}{2\pi n}-\frac{\sin(2\pi na)}{4\pi^2n^2}\right)\end{align}